Lecture 15 Solutions: Measuring the Stars
Practice Problem Solutions
Student note: These are model solutions written to show setup, units, and reasoning clearly. Your own work can be shorter as long as the physics, units, and logic are sound.
Use these after you try the problems yourself. The point is to compare your reasoning to a worked example, not to skip the thinking step.
Core Problems (Master these before the exam)
Problem 1: Parallax and Distance
Restatement: Betelgeuse has parallax \(4.51\) milliarcseconds. We want its distance in parsecs and in light-years.
Given: - \(p = 4.51\) milliarcseconds - \(1\) milliarcsecond \(= 0.001\) arcsecond - \(1\ \text{pc} \approx 3.26\ \text{ly}\)
Find: - (a) \(d\) in parsecs - (b) \(d\) in light-years
Equation: \[ d(\text{pc}) = \frac{1}{p(\text{arcsec})} \]
Solution:
First convert the parallax to arcseconds:
\[ p = 4.51\ \text{mas} = 4.51 \times 10^{-3}\ \text{arcsec} \]
Now compute the distance:
\[ d = \frac{1}{4.51 \times 10^{-3}} \approx 221.7\ \text{pc} \]
Convert parsecs to light-years:
\[ d = 221.7 \times 3.26 \approx 722.7\ \text{ly} \]
Unit check: The parallax formula is defined so that arcseconds map directly to parsecs, so the output unit is parsecs. Multiplying parsecs by \(\text{ly/pc}\) gives light-years.
Sanity check: A parallax of only a few milliarcseconds should correspond to a large distance, so a few hundred parsecs is reasonable.
Answer: Betelgeuse is about \(222\ \text{pc}\), or about \(723\ \text{light-years}\), away.
Problem 2: Inverse-Square Law
Restatement: Two stars have the same luminosity, but one is twice as far away. We want the ratio of their apparent brightnesses.
Key insight: If luminosity is fixed, apparent brightness depends only on distance through the inverse-square law. Doubling the distance makes the star four times dimmer.
Answer: Using
\[ b = \frac{L}{4\pi d^2}, \]
the brightness ratio is
\[ \frac{b_A}{b_B} = \frac{d_B^2}{d_A^2} = \frac{20^2}{10^2} = 4. \]
So Star A appears 4 times brighter than Star B.
Common misconception: Students often think “twice as far means twice as dim.” The inverse-square law says it becomes four times dimmer, not two.
Problem 3: Magnitude Differences
Restatement: Two stars differ by 5 magnitudes. We want the corresponding flux ratio.
Key insight: The magnitude scale is logarithmic, not linear. A difference of 5 magnitudes corresponds to a factor of 100 in brightness.
Answer: The magnitude difference is
\[ \Delta m = m_2 - m_1 = 6.5 - 1.5 = 5. \]
Use the magnitude-flux relation:
\[ \frac{b_1}{b_2} = 10^{\Delta m / 2.5} = 10^{5/2.5} = 10^2 = 100. \]
So Star 1 is 100 times brighter than Star 2.
Common misconception: A larger magnitude means a brighter star. The astronomical magnitude scale runs backward: larger magnitude means fainter.
Problem 4: Luminosity from Distance and Brightness
Restatement: We know the star’s apparent brightness and parallax, and we want its luminosity in watts and in solar luminosities.
Given: - Apparent brightness: \(b = 5 \times 10^{-12}\ \text{W/m}^2\) - Parallax: \(p = 0.2\) arcseconds - \(1\ \text{pc} = 3.086 \times 10^{16}\ \text{m}\) - \(L_\odot = 3.8 \times 10^{26}\ \text{W}\)
Find: \(L\) in watts and in units of \(L_\odot\)
Equation: \[ d(\text{pc}) = \frac{1}{p(\text{arcsec})}, \qquad b = \frac{L}{4\pi d^2} \Rightarrow L = 4\pi d^2 b \]
Solution:
First find the distance:
\[ d = \frac{1}{0.2} = 5\ \text{pc} \]
Convert to meters:
\[ d = 5 \times 3.086 \times 10^{16}\ \text{m} = 1.543 \times 10^{17}\ \text{m} \]
Now compute the luminosity:
\[ L = 4\pi d^2 b \]
\[ L = 4\pi (1.543 \times 10^{17}\ \text{m})^2 (5 \times 10^{-12}\ \text{W/m}^2) \]
\[ L = 4\pi (2.38 \times 10^{34}\ \text{m}^2)(5 \times 10^{-12}\ \text{W/m}^2) \]
\[ L \approx 1.49 \times 10^{24}\ \text{W} \]
Convert to solar luminosities:
\[ \frac{L}{L_\odot} = \frac{1.49 \times 10^{24}}{3.8 \times 10^{26}} \approx 3.9 \times 10^{-3} \]
So
\[ L \approx 0.0039\,L_\odot \]
Unit check: \(\text{m}^2 \times \text{W/m}^2 = \text{W}\), so the luminosity comes out in watts as expected.
Sanity check: This star is nearby but quite faint, so a luminosity much smaller than the Sun’s is reasonable.
Answer: The star’s luminosity is about \(1.5 \times 10^{24}\ \text{W}\), which is about \(0.0039\,L_\odot\).
Problem 5: Distance Modulus
Restatement: A Cepheid’s absolute and apparent magnitudes are given. We want its distance in parsecs and kiloparsecs.
Given: - \(M = -3.5\) - \(m = 13.5\)
Find: \(d\) in parsecs and kiloparsecs
Equation: \[ m - M = 5\log_{10}(d) - 5 \]
Solution:
First compute the distance modulus:
\[ m - M = 13.5 - (-3.5) = 17 \]
Substitute into the equation:
\[ 17 = 5\log_{10}(d) - 5 \]
Add 5 to both sides:
\[ 22 = 5\log_{10}(d) \]
Divide by 5:
\[ \log_{10}(d) = 4.4 \]
Raise 10 to both sides:
\[ d = 10^{4.4} \approx 25{,}119\ \text{pc} \]
Convert to kiloparsecs:
\[ d \approx 25.1\ \text{kpc} \]
Unit check: The distance modulus formula is calibrated so that \(d\) comes out in parsecs.
Sanity check: A bright Cepheid with apparent magnitude \(13.5\) should be quite far away. A distance of tens of kiloparsecs places it in the outer Milky Way, which is plausible.
Answer: The Cepheid is about \(2.5 \times 10^4\ \text{pc}\), or \(25.1\ \text{kpc}\), away.
Challenge Problems (Stretch your thinking)
Challenge 1: The Distance Ladder
Restatement: We observe a Type Ia supernova, use its known absolute magnitude to find the distance, and then compare that distance to Andromeda’s.
Given: - \(M = -19.3\) - \(m = 19.3\) - Andromeda distance \(= 770\ \text{kpc} = 0.77\ \text{Mpc}\)
Find: - (a) \(d\) in megaparsecs - (b) how many times farther than Andromeda
Equation: \[ m - M = 5\log_{10}(d) - 5 \]
Solution:
First compute the distance modulus:
\[ m - M = 19.3 - (-19.3) = 38.6 \]
Substitute:
\[ 38.6 = 5\log_{10}(d) - 5 \]
\[ 43.6 = 5\log_{10}(d) \]
\[ \log_{10}(d) = 8.72 \]
\[ d = 10^{8.72}\ \text{pc} \approx 5.2 \times 10^8\ \text{pc} \]
Convert to megaparsecs:
\[ d \approx 520\ \text{Mpc} \]
Now compare with Andromeda:
\[ \frac{520}{0.77} \approx 675 \]
Unit check: The distance modulus returns parsecs; dividing Mpc by Mpc gives a unitless ratio.
Sanity check: Type Ia supernovae are used precisely because they can be seen at enormous distances, so hundreds of megaparsecs is reasonable.
Answer: The galaxy is about \(520\ \text{Mpc}\) away, which is about 675 times farther than Andromeda.
Challenge 2: Gaia’s Precision
Restatement: We want the distance corresponding to a tiny parallax and the fractional uncertainty in that distance.
Given: - \(p = 10.0 \pm 0.5\) microarcseconds - \(1\ \text{microarcsecond} = 10^{-6}\ \text{arcseconds}\) - \(1\ \text{pc} \approx 3.26\ \text{ly}\)
Find: - (a) \(d\) in parsecs - (b) \(d\) in light-years - (c) fractional uncertainty in \(d\)
Equation: \[ d(\text{pc}) = \frac{1}{p(\text{arcsec})} \]
Solution:
Convert the parallax to arcseconds:
\[ p = 10.0 \times 10^{-6}\ \text{arcsec} = 1.0 \times 10^{-5}\ \text{arcsec} \]
Compute the distance:
\[ d = \frac{1}{1.0 \times 10^{-5}} = 100{,}000\ \text{pc} = 10^5\ \text{pc} \]
Convert to light-years:
\[ d = 10^5 \times 3.26 = 3.26 \times 10^5\ \text{ly} \]
Now compute the fractional parallax uncertainty:
\[ \frac{\delta p}{p} = \frac{0.5 \times 10^{-6}}{10.0 \times 10^{-6}} = 0.05 = 5\% \]
Because \(d = 1/p\), the fractional uncertainty in distance is approximately the same:
\[ \frac{\delta d}{d} \approx 5\% \]
So the absolute distance uncertainty is
\[ \delta d \approx 0.05 \times 100{,}000\ \text{pc} = 5{,}000\ \text{pc} \]
Unit check: A ratio like \(\delta p/p\) is unitless, so the percentage uncertainty is unitless too.
Sanity check: A tiny parallax should correspond to a huge distance, so \(10^5\) pc is sensible.
Answer: The distance is about \(100{,}000\ \text{pc}\) or \(326{,}000\ \text{ly}\), with a fractional uncertainty of \(5\%\).
Challenge 3: An Unseen Companion
Restatement: The total luminosity inferred from the system is larger than the luminosity of the visible star, so we want the missing luminosity.
Key insight: If the observed total light is the sum of two stars, then any mismatch between the total luminosity and the visible star’s luminosity must come from the unseen companion.
Answer: The companion’s luminosity is
\[ L_{\text{companion}} = L_{\text{total}} - L_{\text{primary}} = 2\,L_\odot - 1\,L_\odot = 1\,L_\odot. \]
So the unseen companion has luminosity \(1\,L_\odot\).
In Observable \(\to\) Model \(\to\) Inference form: - Observable: the total flux implies \(2\,L_\odot\) - Model: luminosities in a binary system add - Inference: the unseen companion must contribute the missing \(1\,L_\odot\)
Common misconception: Students sometimes think the companion must be dark because it is “unseen.” Here “unseen” means it is not separately resolved, not that it contributes no light.